Home Forums Trigonometry If $$tanA = \frac{sinB}{1-cosB}$$ find tan2A Reply To: If $$tanA = \frac{sinB}{1-cosB}$$ find tan2A

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Keymaster

To find $$\tan(2A)$$ given that $$\tan(A) = \frac{\sin(B)}{1 – \cos(B)}$$, we’ll first express $$\tan(A)$$ in a more familiar trigonometric identity and then use the double angle formula for $$\tan$$.

Given:

$\tan(A) = \frac{\sin(B)}{1 – \cos(B)}$

Recall the trigonometric identity:

$\tan\left(\frac{\theta}{2}\right) = \frac{\sin(\theta)}{1 + \cos(\theta)}$

However, our expression looks similar but not exactly the same. Notice that for $$\frac{\sin(B)}{1 – \cos(B)}$$, if we consider $$\theta = 2B$$, we get:

$\tan\left(\frac{2B}{2}\right) = \frac{\sin(2B)}{1 + \cos(2B)}$

But our denominator is $$1 – \cos(B)$$, not $$1 + \cos(B)$$. To relate this to the given equation, let’s use the identity:

$\tan(A) = \frac{1 – \cos(\theta)}{\sin(\theta)}$

This is a form of the half-angle identity. However, to directly find $$\tan(2A)$$ from $$\tan(A) = \frac{\sin(B)}{1 – \cos(B)}$$, let’s directly apply the double angle formula for tangent, knowing that $$\tan(A)$$ is given. The double angle formula is:

$\tan(2A) = \frac{2\tan(A)}{1 – \tan^2(A)}$

Substituting $$\tan(A) = \frac{\sin(B)}{1 – \cos(B)}$$ into the formula:

$\tan(2A) = \frac{2\left(\frac{\sin(B)}{1 – \cos(B)}\right)}{1 – \left(\frac{\sin(B)}{1 – \cos(B)}\right)^2}$

Let’s simplify this expression.

After simplifying the expression, we find that:

$\tan(2A) = -\tan(B)$

This result shows that $$\tan(2A)$$ is directly related to $$-\tan(B)$$, given the initial relationship between $$\tan(A)$$ and $$\sin(B)$$ and $$\cos(B)$$.